12
$\begingroup$

Suppose $(X,\tau)$ and $(Y,\sigma)$ are topological spaces. Let $F(X,Y)$ be the set of continuous functions $X\rightarrow Y$. I want to compute the cardinality of $F(X,Y)$. It depends not only on cardinalities $|X|$ and $|Y|$ but on the topologies as well: just imagine what happens if $X$ or $Y$ is discrete.

Question: is it possible to write some "topological invariants" $|\tau|$ and $|\sigma|$ that yield the following explicit formula: $$|F(X,Y)| = \Theta (|X|,|Y|,|\tau|,|\sigma|)$$

I am sorry for being sloppy: I really need to clarify what a formula is. Ideally, $|\tau|$ would be a collection of cardinals and the formula will be given in the cardinal arithmetic. However, I would be glad to have any other method of computing the cardinality. Thus, "the formula" may be something more general.

$\endgroup$
  • 1
    $\begingroup$ There is a continuum that only allow constant mappings and identity to be continuous. May be relevant $\endgroup$ – erz Feb 8 at 20:58
  • 1
    $\begingroup$ If you allow proper class invariants, then the answer is yes, because you can essentially encode the whole of F into the invariant in that case. $\endgroup$ – Joel David Hamkins Feb 8 at 21:25
  • 4
    $\begingroup$ You can also trivially do it if you have global choice, from which you get a definable well-ordering of $V$ of order type $Ord$. For any topological space you find the lowest index $\alpha$ of a homeomorphic topological space, and then your 'cardinal invariant' is $\aleph_\alpha$. $\endgroup$ – James Hanson Feb 8 at 21:32
  • 2
    $\begingroup$ @BugsBunny The point is that a lot of information can be coded in cardinals and under certain common set theoretic assumptions you can actually code the homeomorphism type of $(X,\tau)$ by a unique cardinal. So if you're too permissive about what $|\tau|$ and $\Theta$ are, the answer to your question can become very sensitive to set theoretic assumptions and even trivial under certain specific assumptions. $\endgroup$ – James Hanson Feb 8 at 23:47
  • 1
    $\begingroup$ Next: in <a href="eudml.org/doc/170472"> Wann sind alle stetige abbildungen in $Y$ konstant? </a> Herrlich shows that for every $T_1$-spae $Y$ there is a regular space $X$ such that every continuous map from $X$ to $Y$ is constant, so $|F(X,Y)|=|Y|$. There is no bound on the cardinality of $X$. In particular of $Y$ is the space of rationals $F(X,Y)$ is countable for very large $X$. $\endgroup$ – KP Hart Feb 10 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.